A uniform rod of mass 6M and length 6I is bent to make an equilateral hexagon. If M.I. about an axis passing through the centre of mass and perpendicular to the plane of hexagon is
A
5mI2
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B
6mI2
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C
4mI2
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D
mI2/12
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Solution
The correct option is B5mI2
Clearly, length and mass of each edge =ℓ and M. Consider portion BC : moment of inertia of BC about axis passing its centre and perperdicular to its plane is ml212 .
Thus, moment of inertia about given axis of BC=ml212+md2 (I BC ) (Parallel axis theo. )∴IBC=ml212+m(√3l2)2=ml212+3ml24=56ml2
Hence, total moment of inertia of system =6IBC=5ml2